Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Define momentum and impulse and state their SI units.
• Apply the impulse-momentum theorem to solve problems.
• Explain the conservation of momentum in a closed system.
• Solve problems involving elastic and inelastic collisions.
• Distinguish between elastic and inelastic collisions.

Lesson summary

This week, we delve deeper into the fascinating world of momentum and impulse, building upon the foundational knowledge you gained previously. Understanding momentum and impulse is crucial because it explains how forces affect the motion of objects during collisions. Think about car crashes, the impact of a cricket bat on a ball, or even the recoil of a firearm. These are all real-world scenarios where the principles of momentum and impulse are at play. In South Africa, road safety is a significant concern, making the study of collisions and the forces involved even more relevant.

Lesson notes

2.1 Momentum (p) Momentum is a measure of the "quantity of motion" of an object. It depends on both the mass (m) of the object and its velocity (v).

Definition: Momentum is the product of an object's mass and its velocity.

Formula: `p = mv` SI Unit: kg⋅m/s (kilogram meters per second)

Direction: Momentum is a vector quantity, meaning it has both magnitude and direction. The direction of the momentum is the same as the direction of the velocity.

Example 1: A taxi with a mass of 1500 kg is traveling at a velocity of 20 m/s east. Calculate the taxi's momentum.

Solution: Given: m = 1500 kg, v = 20 m/s east `p = mv = (1500 kg)(20 m/s east) = 30000 kg⋅m/s east` The taxi's momentum is 30000 kg⋅m/s east. 2.2 Impulse (J) Impulse is the change in momentum of an object. It's caused by a force acting on the object over a period of time.

Definition: Impulse is the product of the force acting on an object and the time interval over which it acts. It is also equal to the change in momentum.

Formula: `J = FΔt = Δp = mv_f - mv_i` where `v_f` is the final velocity and `v_i` is the initial velocity.

SI Unit: N⋅s (Newton seconds) or kg⋅m/s (kilogram meters per second) - They are equivalent.

Direction: Impulse is a vector quantity. The direction of the impulse is the same as the direction of the force.

Example 2: A soccer ball with a mass of 0.45 kg is kicked with a force of 120 N for 0.02 seconds. Calculate the impulse imparted to the ball.

Solution: Given: F = 120 N, Δt = 0.02 s `J = FΔt = (120 N)(0.02 s) = 2.4 N⋅s` The impulse imparted to the soccer ball is 2.4 N⋅s. 2.3 Impulse-Momentum Theorem The impulse-momentum theorem states that the impulse acting on an object is equal to the change in momentum of the object. This is a direct consequence of Newton's Second Law.

Formula: `FΔt = Δp = mv_f - mv_i` Example 3: A cricket ball with a mass of 0.16 kg is traveling at 30 m/s towards a batsman. The batsman hits the ball, and it travels in the opposite direction at 40 m/s. Calculate the impulse imparted to the ball. Also, if the contact time between the bat and the ball is 0.005 s, calculate the average force exerted by the bat on the ball.

Solution: Given: m = 0.16 kg, v_i = 30 m/s, v_f = -40 m/s (negative since it's in the opposite direction), Δt = 0.005 s `J = Δp = mv_f - mv_i = (0.16 kg)(-40 m/s) - (0.16 kg)(30 m/s) = -6.4 kg⋅m/s - 4.8 kg⋅m/s = -11.2 kg⋅m/s` The impulse imparted to the ball is -11.2 kg⋅m/s (the negative sign indicates the direction). `FΔt = J` `F = J / Δt = -11.2 kg⋅m/s / 0.005 s = -2240 N` The average force exerted by the bat on the ball is 2240 N (in the opposite direction to the initial velocity of the ball). 2.4 Conservation of Momentum In a closed system (no external forces acting), the total momentum remains constant. This is a fundamental principle in physics.

Formula: `p_initial = p_final` or `m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}` for a system of two objects.

Example 4: Two trolleys are on a frictionless track. Trolley A has a mass of 2 kg and is moving to the right at 3 m/s. Trolley B has a mass of 1 kg and is initially at rest. The trolleys collide and stick together. Calculate the final velocity of the combined trolleys.

Solution: Given: m_A = 2 kg, v_{Ai} = 3 m/s, m_B = 1 kg, v_{Bi} = 0 m/s. After the collision, the trolleys stick together, so v_{Af} = v_{Bf} = v_f. `m_Av_{Ai} + m_Bv_{Bi} = (m_A + m_B)v_f` (Since they stick together, the final mass is the sum of the individual masses) `(2 kg)(3 m/s) + (1 kg)(0 m/s) = (2 kg + 1 kg)v_f` `6 kg⋅m/s = (3 kg)v_f` `v_f = 6 kg⋅m/s / 3 kg = 2 m/s` The final velocity of the combined trolleys is 2 m/s to the right. 2.5 Elastic and Inelastic Collisions Collisions can be classified as either elastic or inelastic based on whether kinetic energy is conserved.

Elastic Collision: A collision in which both momentum and kinetic energy are conserved. In a perfectly elastic collision, no energy is lost as heat or sound. An example is the collision of billiard balls (approximated).

Inelastic Collision: A collision in which momentum is conserved, but kinetic energy is not conserved. Some kinetic energy is transformed into other forms of energy, such as heat, sound, or deformation. Most real-world collisions are inelastic.

Examples: car crashes, a ball of clay hitting a wall. Important

Note: Momentum is ALWAYS conserved in a closed system, regardless of whether the collision is elastic or inelastic. Kinetic energy is only conserved in elastic collisions.

Example 5: Using the scenario from Example 4 (two trolleys colliding and sticking together), determine whether the collision is elastic or inelastic.

Solution: Initial kinetic energy: `KE_i = 1/2 m_A v_{Ai}^2 + 1/2 m_B v_{Bi}^2 = 1/2 (2 kg) (3 m/s)^2 + 1/2 (1 kg) * (0 m/s)^2 = 9 J` Final kinetic energy: `KE_f = 1/2 (m_A + m_B) v_f^2 = 1/2 (3 kg) * (2 m/s)^2 = 6 J` Since `KE_i ≠ KE_f`, the collision is inelastic. Kinetic energy was lost (converted to other forms of energy, likely heat or sound).